62
class is 7, and the interval of these scores is 6. Histogram and polygon are presented in figure 11.
Table 10. Frequency Distribution of data B
2
Class limits
Class boundaries
M idpoint Tally
Frequency Percentage
35 - 41 34.5
– 41.5 38
IIII I 6
54.54 42 - 48
41.5 – 48.5
45 IIII I
6 54.54
49 – 55
48.5 – 55.5
52 IIII I
6 54.54
56 - 62 56.5
– 62.5 59
III 3
27.27 63 - 69
62.5 – 69.5
66 I
1 9.09
22 100
Class Limits
66
Fr e
q u
e n
cy
7 6
59 45
52
Group B Ї
5
38 35
4
1 3
2
Figure 11. Histogram and Polygon of data B
2
B. Normality and Homogeneity
63
The tests that have to be done before analyzing data are normality and homogeneity. The normality test is to check whether the data are in normal
distribution or not. And the homogeneity is applied to find out whether the data are homogeneous or not. This test is important because homogeneity of the data
shows that the population is well-formed. 1. Normality test
If L
o
L obtained is lower than L
t
L table at the level of significance α=0.05 on Liliefors, the sample is in normal distribution. The formula used in
this testing is:
√ ̅
a. Cell A
1
B
1
. In this cell, that contains 11 students having high intelligence who are
taught by using reciprocal teaching n = 11, the highest value of FZ
i
-SF
i
or L
o
is 0.142. L
t
at the level of significance α=0.05 is 0. 249. Because L
o
is lower than L
t
0.142 0.249, it can be concluded that the sample is in normal distribution.
b. Cell A
2
B
1
. In the cell A
2
B
1
, that contains 11 students having high intelligence who are taught by using direct instructional n = 11, the highest value of FZ
i
-SF
i
or L
o
is 0.175. L
t
at the level of significance α=0.05 is 0.249. Because L
o
is
64
lower than L
t
0.175 0.249, it can be concluded that the sample is in normal distribution.
c. Cell A
1
B
2
. In the cell A
1
B
2
, that contains 11 students having low intelligence who are taught by using reciprocal teaching n = 11, the highest value of FZ
i
-SF
i
or L
o
is 0.184. L
t
at the level of significance α=0.05 is 0.249. Because L
o
is lower than L
t
0.184 0.2492, it can be concluded that the sample is in normal distribution.
d. Cell A
2
B
2
. In the cell A
2
B
2
, that contains 11 students having low intelligence who are taught by using direct instructional n = 11, the highest value of FZ
i
-SF
i
or L
o
is 0.171. L
t
at the level of significance α=0.05 is 0.249. Because L
o
is lower than L
t
0.1710.249, it can be concluded that the sample is in normal distribution.
e. Cell A
1
. In the cell A
1
, that contains 22 who are taught by using reciprocal teaching n = 22, the highest value of FZ
i
-SF
i
or L
o
is 0.145. L
t
at the level of significance α=0.05 is 0.190. Because L
o
is lower than L
t
0.145 0.186, it can be concluded that the sample is in normal distribution.
f. Cell A
2
. In the cell A
2
, that contains 22 students who are taught by using direct instructional n = 22, the highest value of FZ
i
-SF
i
or L
o
is 0.096. L
t
at the
65 level of significance α=0.05 is 0.190. Because L
o
is lower than L
t
0.096 0.0.186, it can be concluded that the sample is in normal distribution.
g. Cell B
1
. In the cell B
1
, that contains 22 students who are having high intelligence n = 22, the highest value of FZ
i
-SF
i
or L
o
is 0.151. L
t
at the level of significance α=0.05 is 0.186. Because L
o
is lower than L
t
0.151 0.0.186, it can be concluded that the sample is in normal distribution.
h. Cell B
2
. In the cell B
2
, that contains 22 students who are having low intelligence n = 22, the highest value of FZ
i
-SF
i
or L
o
is 0.146. L
t
at the level of significance α=0.05 is 0.185. Because L
o
is lower than L
t
0.146 0.0.186, it can be concluded that the sample is in normal distribution.
Table 11. The Normality Test
. No Data
The Number of
sample L
Obtained L
o
L Table
L
t
Alfa Distribution
of Population
1 2
3 4
5 6
7 8
A
1
B
1
A
2
B
1
A
1
B
2
A
2
B
2
A
1
A
2
B
1
B
2
11 11
11 11
22 22
22 22
0.142 0.175
0.184 0.171
0.145 0.096
0.151 0.145
0.249 0.249
0.249 0.249
0.186 0.186
0.186 0.186
0.05 0.05
0.05 0.05
0.05 0.05
0.05 0.05
Normal Normal
Normal Normal
Normal Normal
Normal Normal
2. Homogeneity Test
66
The testing is done to find out whether the data are homogeneous. This test is important because homogeneity of the data shows that the population is well-
formed. To test homogeneity of data, test is used. If
2
o
is lower than
2
t
at the level of significance α=0.05, the data are homogeneous.
Table 12. Homogeneity Test
Group n
i
– 1 1 n
i
– 1 s
i 2
Lns
i 2
n
i
– 1 Lns
i 2
1 10
0.1 40.273
3.696 36.957
2 10
0.1 19.473
2.969 29.690
3 10
0.1 57.655
4.054 40.545
4 10
0.1 54.655
4.001 40.010
40 0.4
147.202
2
=
k N
1 1
n 1
1 k
3 1
1 s
ln 1
n B
i 2
i i
=
40 1
4 .
1 4
3 1
1 202
. 147
461 .
150
= 3.128
2 0. 05
= 7.815 Because
2
o
3.128 is lower than 815
. 7
2
t
, the data are homogeneous.
C. Data Analysis
